If x is the sum of six consecutive integers, then x is divis

X= y+(y+1)+(y+2)+(y+3)+(y+4)+(y+5)
= 6y +15
=3(2y+5)

So X is divisible by 3

A. I only

I checked for few more consecutive values of Y...& all proved to be multiples of 3.

So X should be divisible by 3!!

If x is the sum of six consecutive integers, then x is divisible by which of the following:

I. 3
II. 4
III. 6

Let us write the 6 consecutive integers like this ......

a-2 , a-1 , a , a+ 1 , a + 2, a + 3

Sum of these numbers will be 6a + 3

1) Divisible by 3 --> Look at 6a + 3 = 3(2a +1 ) ....Thus it is divisible by 3.

2) Divisible by 4 --> 6a + 3 = 4a + 2a + 3 .... Thus this will only be possible if 2a + 3 is divisble by 3. However 2a + 3 is an odd number thus it is not divisible by 3.

3) Divisible by 6 --> Based on 6a + 3. Dividing this number by 6 will give us remainder of 3. So not divisible by 6.


Thus answer should be A

OA A

OE [spoiler]Since x is the sum of six consecutive integers, it can be written as:

x = n + (n + 1) + (n + 2) + (n + 3) + (n + 4) + (n + 5)
x = 6n + 15

Note that x must be odd since it is the sum of the even term 6n and the odd term 15, and an even plus an odd gives an odd.

I. TRUE: Since 6n and 15 are both divisible by 3, x is divisible by 3.

II. FALSE: Since x is odd, it CANNOT be divisible by 4.

III. FALSE: Since x is odd, it CANNOT be divisible by 6. [/spoiler]

The correct answer is A